Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous

Exercises · 6.67

Q. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.

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Concept understanding — Solubility Product Constant

Solubility Product Constant: From Intuition to Precision

Imagine you drop a pinch of salt into a glass of water. The salt crystals disappear — they dissolve. But what if you keep adding salt, spoonful after spoonful? At some point, the water can't hold any more; the extra salt just sits at the bottom, undissolved. That's a saturated solution — the maximum amount of solute has dissolved at that temperature.

Now, here's the key question: even in that saturated solution, is everything static? Not at all. At the microscopic level, salt ions are constantly leaving the solid crystal and entering the solution (dissolving), while other ions in solution bump into the crystal and stick back (precipitating). At saturation, these two processes happen at exactly the same rate. The system is in dynamic equilibrium.

Note

A saturated solution is not "full" in a static sense — it's a busy, balanced dance between dissolving and precipitating.

The Intuition: A Crowded Dance Floor

Think of a dance hall with a capacity limit. The "dancers" are ions (like Na⁺ and Cl⁻ from table salt). The dance floor is the solution. When the floor is empty, dancers easily find space — dissolution is fast. As more dancers arrive, they start bumping into each other and some leave the floor (precipitate). At the maximum capacity, the number of dancers entering equals the number leaving. That equilibrium number of dancers is what we call solubility.

But here's the twist: for many salts, the "dancers" come in different types — say, positive ions and negative ions. The equilibrium isn't just about the total number; it's about the product of their concentrations. Why product? Because the chance of a positive and a negative ion meeting to form a solid depends on both their concentrations. If you double the concentration of positive ions, the chance of a collision doubles. If you double both, it quadruples.

That product — at equilibrium, for a saturated solution — is a constant. That's the solubility product constant, $K_{sp}$ .

The Precise Statement

For a sparingly soluble salt that dissociates in water as:

A_mB_n(s) \rightleftharpoons m A^{n+}(aq) + n B^{m-}(aq)

the solubility product constant is defined as:

K_{sp} = [A^{n+}]^m \cdot [B^{m-}]^n

where the square brackets denote molar concentrations (mol/L) at saturation (equilibrium with the solid).

Watch out

The solid $A_mB_n$ does not appear in the expression. Its concentration is constant (pure solid) and is absorbed into $K_{sp}$ . Never write $[A_mB_n]$ in the $K_{sp}$ expression.

What $K_{sp}$ Tells You

Small $K_{sp}$ (e.g., $10^{-30}$ ): The salt is very insoluble. Only a tiny amount dissolves.
Large $K_{sp}$ (e.g., $10^{-2}$ ): The salt is relatively soluble.
$K_{sp}$ is temperature-dependent — always quote the temperature (usually 25°C).

Important

$K_{sp}$ is an equilibrium constant. It only applies to saturated solutions in contact with undissolved solid. If no solid is present, the solution may be unsaturated ( $Q < K_{sp}$ ) or supersaturated ( $Q > K_{sp}$ ), but $K_{sp}$ itself doesn't change.

A Concrete Example: Silver Chloride

Silver chloride, AgCl, is a classic sparingly soluble salt. Its dissolution:

AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

The $K_{sp}$ expression:

K_{sp} = [Ag^+][Cl^-]

At 25°C, $K_{sp} = 1.8 \times 10^{-10}$ . This tiny number means that in a saturated solution, the product of the two ion concentrations is only $1.8 \times 10^{-10}$ .

If you know the solubility of AgCl is $s$ mol/L, then $[Ag^+] = s$ and $[Cl^-] = s$ , so: …

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The key idea is the Solubility Product Constant ( $K_{sp}$ ): for a sparingly soluble salt $A_xB_y$ , the product of ion concentrations (raised to stoichiometric coefficients) at saturation is constant at a given temperature.

General method (for a salt $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$ ):

Let molar solubility = $s$ mol/L.
Then $[A^{y+}] = xs$ , $[B^{x-}] = ys$ .
$K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y}$ .
Solve for $s$ , then compute individual ion molarities.

Using standard $K_{sp}$ values at 298 K (from Table 6.9, NCERT):

Salt	$K_{sp}$	$x$	$y$	$s$ (mol/L)	$[cation]$	$[anion]$
$\text{Ag}_2\text{CrO}_4$	$1.1 \times 10^{-12}$	2	1	$s = \sqrt[3]{\frac{K_{sp}}{4}} = 6.5 \times 10^{-5}$	$1.3 \times 10^{-4}$	$6.5 \times 10^{-5}$
$\text{BaCrO}_4$	$1.2 \times 10^{-10}$	1	1	$s = \sqrt{K_{sp}} = 1.1 \times 10^{-5}$	$1.1 \times 10^{-5}$	$1.1 \times 10^{-5}$

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The solubility of a sparingly soluble salt is found by relating its $K_{sp}$ expression to the stoichiometric concentrations of its ions. For each salt, we set up the dissolution equilibrium, let $s$ be the molar solubility, substitute into the $K_{sp}$ formula, and solve for $s$ . The individual ion molarities then follow from the stoichiometric coefficients. The results are tabulated below.

The key idea is that the solubility product constant $K_{sp}$ is the equilibrium constant for the dissolution of a sparingly soluble salt. It is the product of the concentrations of the ions, each raised to the power of its stoichiometric coefficient in the balanced equation. For a salt $A_xB_y$ that dissolves as:

A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)

the $K_{sp}$ expression is:

K_{sp} = [A^{y+}]^x [B^{x-}]^y

If we let the molar solubility be $s$ mol/L (the number of moles of salt that dissolve per litre of solution), then from the stoichiometry:

[A^{y+}] = x s \quad \text{and} \quad [B^{x-}] = y s

Substituting into the $K_{sp}$ expression gives:

K_{sp} = (x s)^x (y s)^y = x^x y^y s^{x+y}

We then solve for $s$ . The molarities of the individual ions are then $x s$ and $y s$ respectively.

Now, we apply this to each salt. The $K_{sp}$ values at 298 K are taken from Table 6.9 (standard NCERT data). Let's work through each one.

1. Silver chromate, $Ag_2CrO_4$

Dissociation: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$

Here, $x = 2$ , $y = 1$ . Let solubility = $s$ mol/L.

Then $[Ag^+] = 2s$ , $[CrO_4^{2-}] = s$ .

K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2s)^2 (s) = 4s^3

From Table 6.9, $K_{sp}(Ag_2CrO_4) = 1.1 \times 10^{-12}$ .

4s^3 = 1.1 \times 10^{-12} \implies s^3 = \frac{1.1 \times 10^{-12}}{4} = 2.75 \times 10^{-13}

s = \sqrt[3]{2.75 \times 10^{-13}} = \sqrt[3]{275 \times 10^{-15}} = \sqrt[3]{275} \times 10^{-5}

Since $\sqrt[3]{275} \approx 6.5$ (because $6.5^3 = 274.6$ ), we get:

s \approx 6.5 \times 10^{-5} \text{ mol/L}

Ion molarities: $[Ag^+] = 2s = 1.3 \times 10^{-4}$ M, $[CrO_4^{2-}] = s = 6.5 \times 10^{-5}$ M.

Watch out

A common mistake is to forget the coefficient 2 on $Ag^+$ when squaring. The $K_{sp}$ is $(2s)^2(s) = 4s^3$ , not $s^3$ . Always write the full expression from the balanced equation.

2. Barium chromate, $BaCrO_4$

Dissociation: $BaCrO_4(s) \rightleftharpoons Ba^{2+}(aq) + CrO_4^{2-}(aq)$

Here, $x = 1$ , $y = 1$ . Let solubility = $s$ mol/L.

Then $[Ba^{2+}] = s$ , $[CrO_4^{2-}] = s$ .

K_{sp} = [Ba^{2+}][CrO_4^{2-}] = s \cdot s = s^2

From Table 6.9, $K_{sp}(BaCrO_4) = 1.2 \times 10^{-10}$ .

s^2 = 1.2 \times 10^{-10} \implies s = \sqrt{1.2 \times 10^{-10}} = \sqrt{1.2} \times 10^{-5}

Since $\sqrt{1.2} \approx 1.095$ , we get:

s \approx 1.1 \times 10^{-5} \text{ mol/L}

Ion molarities: $[Ba^{2+}] = s = 1.1 \times 10^{-5}$ M, $[CrO_4^{2-}] = s = 1.1 \times 10^{-5}$ M.

Tip

For a 1:1 salt like $BaCrO_4$ , the solubility is simply $\sqrt{K_{sp}}$ . This is the simplest case.

3. Ferric hydroxide, $Fe(OH)_3$

Dissociation: $Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq)$

Here, $x = 1$ , $y = 3$ . Let solubility = $s$ mol/L.

Then $[Fe^{3+}] = s$ , $[OH^-] = 3s$ .

K_{sp} = [Fe^{3+}][OH^-]^3 = (s)(3s)^3 = s \cdot 27 s^3 = 27 s^4

From Table 6.9, $K_{sp}(Fe(OH)_3) = 1.0 \times 10^{-38}$ .

27 s^4 = 1.0 \times 10^{-38} \implies s^4 = \frac{1.0 \times 10^{-38}}{27} \approx 3.70 \times 10^{-40}

s = \sqrt[4]{3.70 \times 10^{-40}} = \sqrt[4]{3.70} \times 10^{-10}

Since $\sqrt[4]{3.70} \approx 1.39$ (because $1.4^4 = 3.84$ , close enough), we get:

s \approx 1.39 \times 10^{-10} \text{ mol/L}

Ion molarities: $[Fe^{3+}] = s = 1.39 \times 10^{-10}$ M, $[OH^-] = 3s = 4.17 \times 10^{-10}$ M.

Note

The exponent on $s$ is $x+y = 1+3 = 4$ , so we take the fourth root. The very small $K_{sp}$ reflects the extreme insolubility of $Fe(OH)_3$ .

4. Lead chloride, $PbCl_2$

Dissociation: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$

Here, $x = 1$ , $y = 2$ . Let solubility = $s$ mol/L.

Then $[Pb^{2+}] = s$ , $[Cl^-] = 2s$ .

K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = s \cdot 4 s^2 = 4 s^3

From Table 6.9, $K_{sp}(PbCl_2) = 1.6 \times 10^{-5}$ .

4 s^3 = 1.6 \times 10^{-5} \implies s^3 = \frac{1.6 \times 10^{-5}}{4} = 4.0 \times 10^{-6}

s = \sqrt[3]{4.0 \times 10^{-6}} = \sqrt[3]{4.0} \times 10^{-2}

Since $\sqrt[3]{4.0} \approx 1.587$ , we get:

s \approx 1.59 \times 10^{-2} \text{ mol/L}

Ion molarities: $[Pb^{2+}] = s = 1.59 \times 10^{-2}$ M, $[Cl^-] = 2s = 3.18 \times 10^{-2}$ M.

Watch out

$PbCl_2$ has a relatively high $K_{sp}$ compared to the others, so its solubility is in the $10^{-2}$ M range — it is not "insoluble" in the strict sense, but sparingly soluble. Always check the magnitude.

5. Mercurous iodide, $Hg_2I_2$

Dissociation: $Hg_2I_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2I^-(aq)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{AgBrO}_3$
›Reveal solutionSolution

$s = \sqrt{K_{sp}} = 3.05\times10^{-5}\ \text{mol L}^{-1}$ ; in 100 mL this is $7.20\times10^{-4}\ \text{g}$ of $\text{AgBrO}_3$ — option (D).

Dissolution equilibrium
$\text{AgBrO}_3(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{BrO}_3^{-}(aq), \qquad K_{sp} = s^{2}.$
Molar solubility
$s = \sqrt{9.3\times10^{-10}} = 3.05\times10^{-5}\ \text{mol L}^{-1}.$
Mass in 100 mL (0.1 L)

$n = s \times 0.1 = 3.05\times10^{-6}\ \text{mol},$ …
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KCET 2024Set B-21 markMCQ
$\mathrm{CaC_2O_4}$
›Reveal solutionSolution
For a sparingly soluble salt like $\mathrm{CaC_2O_4}$ that dissociates into two ions, solubility $s$ is the square root of $K_{sp}$ . Here $s = \sqrt{4 \times 10^{-9}} = 2 \times 10^{-4.5} = 6.3 \times 10^{-5}$ mol L $^{-1}$ , so the answer is (A).

The key idea is that solubility and solubility product are linked by the stoichiometry of dissociation. For a salt like $\mathrm{CaC_2O_4}$ , which breaks into one $\mathrm{Ca}^{2+}$ and one $\mathrm{C}_2\mathrm{O}_4^{2-}$ ion, the relationship is especially simple: if $s$ is the solubility in mol L $^{-1}$ , then at saturation $[\mathrm{Ca}^{2+}] = s$ and $[\mathrm{C}_2\mathrm{O}_4^{2-}] = s$ , so $K_{sp} = s \cdot s = s^2$ .

This means you don't need to set up an ICE table or worry about common ions — it's a direct square root. The trap many students fall into is forgetting that $K_{sp}$ is given in units of $(\text{mol L}^{-1})^2$ , which already tells you it's the product of two concentrations, each equal to $s$ .

Write the dissociation equilibrium:

$\mathrm{CaC_2O_4(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + \mathrm{C_2O_4^{2-}(aq)}$

Let the solubility be $s$ mol L $^{-1}$ . Then:

$[\mathrm{Ca}^{2+}] = s$ , $[\mathrm{C}_2\mathrm{O}_4^{2-}] = s$

The solubility product expression is:

$K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{C}_2\mathrm{O}_4^{2-}] = s \cdot s = s^2$

Substitute the given value:

$s^2 = 4 \times 10^{-9}$

Take the square root:

$s = \sqrt{4 \times 10^{-9}} = \sqrt{4} \times \sqrt{10^{-9}} = 2 \times 10^{-4.5}$ …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{A}_3 \mathrm{~B}_4$
›Reveal solutionSolution
The key is to convert the given solubility in g/L to molar solubility (mol/L), then write the Ksp expression for the salt A₃B₄ in terms of that molar solubility. The result is $K_{sp} = 6912 \, (s/M)^7$ , which corresponds to option (A).

Concept & Intuition

For a sparingly soluble salt like A₃B₄, the dissolution equilibrium is:
$\mathrm{A_3B_4(s)} \rightleftharpoons 3\mathrm{A^{n+}(aq)} + 4\mathrm{B^{m-}(aq)}$
The solubility product $K_{sp}$ is the product of the ion concentrations at equilibrium, each raised to the power of its stoichiometric coefficient. If we know the molar solubility (mol/L), we can directly find the ion concentrations. Here, solubility is given in g/L, so we first convert to molar solubility using the molar mass M.

Step-by-step solution

Convert solubility from g/L to mol/L

Solubility in g/L = $s$ . Molar mass = M g/mol.

Molar solubility (mol/L) = $\frac{s}{M}$ .

Write the dissolution equilibrium

$\mathrm{A_3B_4(s)} \rightleftharpoons 3\mathrm{A^{n+}} + 4\mathrm{B^{m-}}$
If the molar solubility is $x = s/M$ , then:
$[\mathrm{A^{n+}}] = 3x, \quad [\mathrm{B^{m-}}] = 4x$

Write the Ksp expression

$K_{sp} = [\mathrm{A^{n+}}]^3 [\mathrm{B^{m-}}]^4$
Substitute the concentrations:
$K_{sp} = (3x)^3 \cdot (4x)^4$

Simplify the expression

$(3x)^3 = 27x^3, \quad (4x)^4 = 256x^4$
Multiply:
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KCET 2023Set D-21 markMCQ
$1a : \sqrt{3}a : \sqrt{2}a$
›Reveal solutionSolution

In each cubic lattice, find the direction along which the spheres actually touch, count how many radii lie along it, and equate to the length of that line in terms of $a$ .

1. Simple cubic (SC)

Atoms sit only at the corners and touch along the cube edge. The edge of length $a$ contains two half-atoms $\Rightarrow$ 2 radii:
$2r = a \quad\Longrightarrow\quad \boxed{r_{SC} = \frac{a}{2}}$
2. Body-centred cubic (BCC)

The corner atoms do not touch each other; the contact is corner–body-centre–corner, i.e. along the body diagonal, whose length is $\sqrt{3}\,a$ . That diagonal contains 4 radii ( $r + 2r + r$ ):
$4r = \sqrt{3}\,a \quad\Longrightarrow\quad \boxed{r_{BCC} = \frac{\sqrt{3}}{4}a}$
3. Face-centred cubic (FCC)

Here the contact is corner–face-centre–corner, i.e. along the face diagonal, of length $\sqrt{2}\,a$ , which contains 4 radii:

$4r = \sqrt{2}\,a \quad\Longrightarrow\quad r_{FCC} = \frac{\sqrt{2}}{4}a = \boxed{\frac{a}{2\sqrt{2}}}$ …
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KCET 2023Set D-21 markMCQ
$_3$
›Reveal solutionSolution

Both routes give the same hydrated ferric oxide sol; what differs is the peptising ion adsorbed — $\mathrm{Fe^{3+}}$ from hot-water hydrolysis (positive sol) versus $\mathrm{OH^-}$ from the alkaline medium (negative sol).

1. The dispersed phase in both cases

Both reactions produce the same colloidal particle — hydrated ferric oxide:
$\mathrm{FeCl_3} + 3\mathrm{H_2O} \;\xrightarrow{\text{hot}}\; \mathrm{Fe(OH)_3 / Fe_2O_3\cdot xH_2O} + 3\mathrm{HCl}$
So every option has the same dispersed phase; the discriminator is the charge-conferring (adsorbed) ion, written after the slash.

2. The rule that decides the charge — preferential adsorption

A colloidal particle preferentially adsorbs the ion common to its own lattice that is present in the medium (Hardy–Schulze / preferential-adsorption idea).

Sol X — $\mathrm{FeCl_3}$ added to excess HOT WATER:

The medium is rich in $\mathrm{Fe^{3+}}$ (from the ferric chloride itself). The $\mathrm{Fe_2O_3\cdot xH_2O}$ particles adsorb $\mathrm{Fe^{3+}}$ ions, which are common to their lattice.
$\Rightarrow \mathbf{X} = \mathrm{Fe_2O_3\cdot xH_2O / Fe^{3+}} \quad\text{— a \textbf{positively} charged sol}$
Sol Y — $\mathrm{FeCl_3}$ added to NaOH(aq):

Now the medium is alkaline, full of $\mathrm{OH^-}$ , which is the ion common to $\mathrm{Fe(OH)_3}$ . The particles adsorb $\mathrm{OH^-}$ .
$\Rightarrow \mathbf{Y} = \mathrm{Fe_2O_3\cdot xH_2O / OH^{-}} \quad\text{— a \textbf{negatively} charged sol}$
3. Why this matters …
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COMEDK 2023Set 2023-E1 markMCQ
$0.2 \mathrm{~g}$
›Reveal solutionSolution

Volume of a saturated solution that holds this many moles: V = n / s = 7.194 x 10^-4 / 2 x 10^-3 = 0.3597 L = 359.7 mL

Concept: solubility from Ksp of an AB2-type salt, then volume = moles / solubility.

PbCl2 -> Pb2+ + 2Cl-

Ksp = (s)(2s)^2 = 4 s^3

4 s^3 = 3.2 x 10^-8

s^3 = 8 x 10^-9

s = 2 x 10^-3 mol/L

Moles of PbCl2 to be dissolved:

n = 0.2 / 278 = 7.194 x 10^-4 mol …
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KCET 2019Set A-11 markMCQ
$1.5 \times 10^{-4}$
›Reveal solutionSolution
Micelles form only above the critical micelle concentration (CMC). Since the CMC is $1.5 \times 10^{-4}$ mol L $^{-1}$ , the only concentration above it is $2.0 \times 10^{-3}$ mol L $^{-1}$ . The correct option is (A).

The key idea is simple: micelle formation is not spontaneous at any concentration. Soap molecules (surfactants) exist as individual ions or molecules in dilute solution. As concentration increases, they eventually reach a threshold called the critical micelle concentration (CMC). Above this value, the molecules cluster into micelles — spherical aggregates with hydrophobic tails inward and hydrophilic heads outward. Below the CMC, no micelles form.

So the question reduces to: which of the given concentrations is greater than the CMC of $1.5 \times 10^{-4}$ mol L $^{-1}$ ?

Compare each option to the CMC:

(A) $2.0 \times 10^{-3}$ — this is $20 \times 10^{-4}$ , clearly larger than $1.5 \times 10^{-4}$ .

(B) $7.5 \times 10^{-5}$ — this is $0.75 \times 10^{-4}$ , smaller than the CMC.

(C) $4.6 \times 10^{-5}$ — this is $0.46 \times 10^{-4}$ , also smaller.

(D) $1.1 \times 10^{-4}$ — this is $1.1 \times 10^{-4}$ , still less than $1.5 \times 10^{-4}$ .

Only option (A) exceeds the CMC. Therefore, micelle formation is possible only at that concentration. …
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Q.Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.

1. Silver chromate, Ag2CrO4Ag_2CrO_4Ag2​CrO4​

2. Barium chromate, BaCrO4BaCrO_4BaCrO4​

3. Ferric hydroxide, Fe(OH)3Fe(OH)_3Fe(OH)3​

4. Lead chloride, PbCl2PbCl_2PbCl2​

5. Mercurous iodide, Hg2I2Hg_2I_2Hg2​I2​

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1. Silver chromate, $Ag_2CrO_4$

2. Barium chromate, $BaCrO_4$

3. Ferric hydroxide, $Fe(OH)_3$

4. Lead chloride, $PbCl_2$

5. Mercurous iodide, $Hg_2I_2$

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